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insens350 [35]
3 years ago
8

Can someone help me please

Chemistry
1 answer:
MArishka [77]3 years ago
3 0

Answer:

Group 4A (or IVA) of the periodic table includes the nonmetal carbon (C), the metalloids silicon (Si) and germanium (Ge), the metals tin (Sn) and lead (Pb), and the yet-unnamed artificially-produced element ununquadium (Uuq).

The Group 4A elements have four valence electrons in their highest-energy orbitals (ns2np2). Carbon and silicon can form ionic compounds by gaining four electrons, forming the carbide anion (C4-) and silicide anion (Si4-), but they more frequently form compounds through covalent bonding. Tin and lead can lose either their outermost p electrons to form 2+ charges (Sn2+, the stannous ion, and Pb2+, the plumbous ion) or their outermost s and p electrons to form 4+ charges (Sn4+, the stannic ion, and Pb4+, the plumbic ion).

Carbon (C, Z=6).

Carbon is most familiar as a black solid is graphite, coal, and charcoal, or as the hard, crystalline diamond form. The name is derived from the Latin word for charcoal, carbo. It is found in the Earth's crust at a concentration of 480 ppm, making it the 15th most abundant element. It is found in form of calcium carbonate, CaCO3, in minerals such as limestone, marble, and dolomite (a mixture of calcium and

Explanation:

<em><u>T</u></em><em><u>H</u></em><em><u>I</u></em><em><u>S</u></em><em><u> </u></em><em><u>A</u></em><em><u>L</u></em><em><u>L</u></em><em><u> </u></em><em><u>I</u></em><em><u> </u></em><em><u>K</u></em><em><u>N</u></em><em><u>O</u></em><em><u>W</u></em>

<u>E</u><u>N</u><u>J</u><u>O</u><u>Y</u><u> </u><u>THE</u><em><u> </u></em><em><u>A</u></em><em><u>N</u></em><em><u>S</u></em><em><u>W</u></em><em><u>E</u></em><em><u>R</u></em>

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Answer:

\boxed{\text{10 L}}

Explanation:

We have two pressures, two temperatures, and one volume.

This looks like a question in which we can use the Combined Gas Law to calculate the volume.

\dfrac{p_{1}V_{1}}{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}

Data:

\begin{array}{rclrclrcl}p_{1}& =& \text{2 atm}\qquad & V_{1} &= & \text{10 L}\qquad & T_{2}& =& \text{400 K}\\p_{2}& =& \text{4 atm}\qquad & V_{2} &= & \text{?}\qquad & T_{2}& =& \text{800 K}\\\end{array}

Calculation:

\begin{array}{rcl}\dfrac{2 \times 10}{400}& =& \dfrac{4V_{2} }{800}\\\\0.050& = &0.0050V_{2}\\V_{2}& = &\mathbf{10 L}\end{array}\\\text{The final volume is }\boxed{\textbf{10 L}}

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A. 0.83

Explanation:

Hopefully this helps!

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A 5.5 g sample of a substance contains only carbon and oxygen. Carbon makes up 35% of the mass of the substance. The rest is mad
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We have been given the condition that carbon makes up 35% of the mass of the substance and the rest is made up of oxygen. With this, it can be concluded that 65% of the substance is made up of oxygen. If we let x be the mass of oxygen in the substance, the operation that would best represent the scenario is,

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